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I am trying to evaluate the definite integral of (ex. 36 pg. 523, James Stewart Calculus 7e): $$\int_{-\frac {\pi}{4}}^{\frac {\pi}{4}} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx$$ However, I have not got any possible approach recently. I have tried using trigonometric substitution but $x^2$ keep challenging me. I would be grateful if there is any suggestions, thank you!

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    $\begingroup$ The integrand is an odd function. $\endgroup$ Sep 6 at 9:09
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Take the function, I will call it $f$, and plug in $-x$ for $x$. What do you get? You get exactly $-f(x)$. Now notice that you are integrating over the interval $(-\frac{\pi}{4},\frac{\pi}{4})$. That which you will add in the positive region, you will substract by integrating in the negative region. Hence the result is $0$.

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$I=\displaystyle\int_{-\frac {\pi}{4}}^{\frac {\pi}{4}} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx$

$I=\displaystyle\int_{-\frac {\pi}{4}}^{0} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx+\int_{0}^{\frac {\pi}{4}} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx=-\int_{0}^{-\frac {\pi}{4}} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx+\int_{0}^{\frac {\pi}{4}} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx$

Let $u=-x\implies du=-dx$

$\displaystyle I=-\int_{0}^{\frac {\pi}{4}} \frac {x^2 \tan (-u)}{1+{\cos^4{-u}}}(-du)+\int_{0}^{\frac {\pi}{4}} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx$

$\displaystyle I=-\int_{0}^{\frac {\pi}{4}} \frac {x^2 \tan (u)}{1+{\cos^4{u}}}(du)+\int_{0}^{\frac {\pi}{4}} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx$

$\displaystyle I=-\int_{0}^{\frac {\pi}{4}} \frac {x^2 \tan (x)}{1+{\cos^4{x}}}(dx)+\int_{0}^{\frac {\pi}{4}} \frac {x^2 \tan x}{1+{\cos^4{x}}}dx$

$I=0$

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The integrad is an odd function: $f(-x)=-f(x)$, so $$\int_{-\pi/4}^{\pi/4}f(x) dx=0.$$

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    $\begingroup$ Did you check the comment? $\endgroup$
    – Ilovemath
    Sep 6 at 10:10
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The function $f(x)=\frac{x^2 \tan x}{1+\cos^4 x}$ is odd. Hence any integral of the form $\int_{-a}^{a} {f(x)} {dx} $ is equal to zero.

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